Event A has a probability of occurring of 4/5 . Event B has a probability of occurring of 1/3. If events A and B are independent
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We will use the right Riemann sum. We can break this integral in two parts.
We take the interval and we divide it n times:
The area of the i-th rectangle in the right Riemann sum is:
For the first part of our integral we have:
For the second part we have:
We can now put it all together:
![\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\ \sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]](https://tex.z-dn.net/?f=%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%20%5B%28%5CDelta%20x%29%5E4%20i%5E3-6%28%5CDelta%20x%29%5E2i%5D%5C%5C%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%5B%20%28%5Cfrac%7B3%7D%7Bn%7D%29%5E4%20i%5E3-6%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5D%5C%5C%0A%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5B%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2%20i%5E2-6%5D)
We can also write n-th partial sum:
We take the interval and we divide it n times:
The area of the i-th rectangle in the right Riemann sum is:
For the first part of our integral we have:
For the second part we have:
We can now put it all together:
We can also write n-th partial sum: